Consider two lines: line $l$ parametrized as
\begin{align*} 
x &= 1 + 4t,\\
y &=  4 + 3t
\end{align*}and the line $m$ parametrized as
\begin{align*} 
x &=-5  + 4s\\
y &=  6 + 3s.
\end{align*}Let $A$ be a point on line $l$, $B$ be a point on line $m$, and let $P$ be the foot of the perpendicular from $A$ to line $m$.

Then $\overrightarrow{PA}$ is the projection of $\overrightarrow{BA}$ onto some vector $\begin{pmatrix} v_1\\v_2\end{pmatrix}$ such that $v_1+v_2 = 2$. Find $\begin{pmatrix}v_1 \\ v_2 \end{pmatrix}$.
Solution: As usual, we start by graphing these lines. An easy way to go about it is to plot some points. Let's plug in $t =0$ and $t = 1$ for line $l$, getting the points $(1, 4)$ and $(5, 7)$. Here's our line:

[asy]
size(200);
import TrigMacros;
import olympiad;

//Gives the maximum line that fits in the box. 
path maxLine(pair A, pair B, real xmin, real xmax, real ymin, real ymax) 
{
    path[] endpoints; 
    endpoints = intersectionpoints(A+10(B-A) -- A-10(B-A), (xmin, ymin)--(xmin, ymax)--(xmax, ymax)--(xmax, ymin)--cycle); 
    return endpoints[1]--endpoints[0]; 
}

pair A= (1,4); 
pair B = (-5, 6);

//Direction vector of the parallel lines
pair dir = (4,3);

//Foot of the perpendicular from A to the other line
pair P = foot(A, B-dir, B+dir);

rr_cartesian_axes(-8,8,-5,12,complexplane=false,usegrid=true);

draw(maxLine(A,A+dir, -8,8,-5,12)); 

label("$l$", A-1.8dir, SE);

dot("$t = 0$", A, SE);
dot("$t = 1$", A + dir, SE); 

[/asy]
Similarly, we plug in $s = 0$ and $s = 1$ for line $m$, getting the points $(-5, 6)$ and $(-1, 9)$:

[asy]
size(200);
import TrigMacros;
import olympiad;

//Gives the maximum line that fits in the box. 
path maxLine(pair A, pair B, real xmin, real xmax, real ymin, real ymax) 
{
    path[] endpoints; 
    endpoints = intersectionpoints(A+10(B-A) -- A-10(B-A), (xmin, ymin)--(xmin, ymax)--(xmax, ymax)--(xmax, ymin)--cycle); 
    return endpoints[1]--endpoints[0]; 
}

pair A = (1,4); 
pair B = (-5, 6);


//Direction vector of the parallel lines
pair dir = (4,3);

//Foot of the perpendicular from A to the other line
pair P = foot(A, B-dir, B+dir);

rr_cartesian_axes(-8,8,-5,12,complexplane=false,usegrid=true);

draw(maxLine(A,A+dir, -8,8,-5,12)); 
draw(maxLine(B,B+dir, -8,8,-5,12)); 

label("$l$", A+dir, SE); 
label("$m$",P+dir, NW); 

dot("$s = 0$", B, NW);
dot("$s = 1$", B + dir,NW); 

[/asy]

Now we label some points $A$ and $B$, as well as point $P$, and we draw in our vectors:

[asy]
size(200);
import TrigMacros;
import olympiad;

//Gives the maximum line that fits in the box. 
path maxLine(pair A, pair B, real xmin, real xmax, real ymin, real ymax) 
{
    path[] endpoints; 
    endpoints = intersectionpoints(A+10(B-A) -- A-10(B-A), (xmin, ymin)--(xmin, ymax)--(xmax, ymax)--(xmax, ymin)--cycle); 
    return endpoints[1]--endpoints[0]; 
}

pair A = (1,4);
pair B= (-5, 6); 

//Direction vector of the parallel lines
pair dir = (4,3);

//Foot of the perpendicular from A to the other line
pair P = foot(A, B-dir, B+dir);

rr_cartesian_axes(-8,8,-5,12,complexplane=false,usegrid=true);

draw(maxLine(A,A+dir, -8,8,-5,12)); 
draw(maxLine(B,B+dir, -8,8,-5,12));
draw(P--A, red, Arrow(size = 0.3cm)); 
draw(B--A, blue, Arrow(size = 0.3cm)); 
draw(rightanglemark(A, P, P + (P-B), 15));

label("$l$", A+dir, SE); 
label("$m$", P+dir, NW); 

dot("$A$", A, SE);
dot("$P$", P, NW);
dot("$B$", B, NW);

[/asy]
Recall that when we project $\mathbf{v}$ onto $\mathbf{u}$, we place the tail of $\mathbf{v}$ onto a line with direction $\mathbf{u}$, then we drop a perpendicular and draw the vector from the tail of $\mathbf{v}$ to the foot of the perpendicular.

This picture actually doesn't look like our usual projection picture! The vector we're projecting and the projection aren't tail to tail, which makes things harder to visualize. Let's shift the vector over and see if it helps, choosing $Q$ such that
\[\overrightarrow{BQ} = \overrightarrow{PA}.\]Here's the picture:

[asy]
size(200);
import TrigMacros;
import olympiad;

//Gives the maximum line that fits in the box. 
path maxLine(pair A, pair B, real xmin, real xmax, real ymin, real ymax) 
{
    path[] endpoints; 
    endpoints = intersectionpoints(A+10(B-A) -- A-10(B-A), (xmin, ymin)--(xmin, ymax)--(xmax, ymax)--(xmax, ymin)--cycle); 
    return endpoints[1]--endpoints[0]; 
}

pair A = (1,4);
pair B= (-5, 6); 

//Direction vector of the parallel lines
pair dir = (4,3);

//Foot of the perpendicular from A to the other line
pair P = foot(A, B-dir, B+dir);

//End of the shifted vector PA: 
pair Q = B+A-P; 

rr_cartesian_axes(-8,8,-5,12,complexplane=false,usegrid=true);

draw(maxLine(A,A+dir, -8,8,-5,12)); 
draw(maxLine(B,B+dir, -8,8,-5,12));
draw(P--A, red, Arrow(size = 0.3cm)); 
draw(B--A, blue, Arrow(size = 0.3cm)); 
draw(rightanglemark(A, P, P + (P-B), 15));
draw(B--Q, red, Arrow(size = 0.3cm)); 
draw(rightanglemark(B,Q, A-2*dir, 15));

label("$l$", A+dir, SE); 
label("$m$", P+dir, NW); 

dot("$A$", A, SE);
dot("$P$", P, NW);
dot("$Q$",Q, SE);
dot("$B$", B, NW);

[/asy]
That looks better! Our shifted vector $\overrightarrow{BQ}$ is tail to tail with the vector being projected. In fact, since this vector is perpendicular to lines $l$ and $m$, we know that it lies along a line with direction
\[\mathbf{u} = \begin{pmatrix} 3 \\-4 \end{pmatrix}.\]Here's the picture with the line added in:


[asy]
size(200);
import TrigMacros;
import olympiad;

//Gives the maximum line that fits in the box. 
path maxLine(pair A, pair B, real xmin, real xmax, real ymin, real ymax) 
{
    path[] endpoints; 
    endpoints = intersectionpoints(A+10(B-A) -- A-10(B-A), (xmin, ymin)--(xmin, ymax)--(xmax, ymax)--(xmax, ymin)--cycle); 
    return endpoints[1]--endpoints[0]; 
}

pair A = (1,4);
pair B= (-5, 6); 

//Direction vector of the parallel lines
pair dir = (4,3);

//Foot of the perpendicular from A to the other line
pair P = foot(A, B-dir, B+dir);

//End of the shifted vector PA: 
pair Q = B+A-P; 

rr_cartesian_axes(-8,8,-5,12,complexplane=false,usegrid=true);

draw(maxLine(A,A+dir, -8,8,-5,12)); 
draw(maxLine(B,B+dir, -8,8,-5,12));
draw(maxLine(B,Q, -8,8,-5,12));

draw(P--A, red, Arrow(size = 0.3cm)); 
draw(B--A, blue, Arrow(size = 0.3cm)); 
draw(rightanglemark(A, P, P + (P-B), 15));
draw(B--Q, red, Arrow(size = 0.3cm)); 
draw(rightanglemark(B,Q, A-2*dir, 15));

label("$l$", A+dir, SE); 
label("$m$", P+dir, NW); 

dot("$A$", A, SE);
dot("$P$", P, NW);
dot("$Q$",Q, 2*S);
dot("$B$", B, 2*S);

[/asy]

If you want to make sure you're visualizing this correctly, imagine the picture above with lines $l$ and $m$ removed: it should become clear that
\[\overrightarrow{BQ}  = \text{The projection of $\overrightarrow{BA}$ onto } \begin{pmatrix} 3 \\-4 \end{pmatrix}.\]Of course, since $\overrightarrow{PA}$ is equal to $\overrightarrow{BQ}$, we see that
\[\overrightarrow{PA}  = \text{The projection of $\overrightarrow{BA}$ onto } \begin{pmatrix} 3 \\-4 \end{pmatrix}.\]Now, we need to be projecting onto a vector whose components add to $2$. We know that we're in fact projecting onto any non-zero scalar multiple of our vector, so we use
\[-2\mathbf{u} = \begin{pmatrix} -6 \\ 8 \end{pmatrix}\]instead. Therefore, $\overrightarrow{PA}$ is the projection of $\overrightarrow{BA}$ onto $\boxed{\begin{pmatrix}-6 \\ 8 \end{pmatrix}}.$